Euclidean Space and its Isometry GroupBarry Monson – U. N. B.

A. Introduction and References.

There are numerous sources which cover the material below. Many of the prob-lems concerning regular polygons and polyhedra are substantially adapted from thebeautiful treatment in

[1] H. S. M Coxeter, Regular Complex Polytopes, Cambridge University Press,Cambridge, 1991, chapters 1–3.

For a different development of the isometry groups and regular polyhedra in var-ious spaces, consult

[2] L. Fejes-Toth, Regular Figures, Pergammon, New York, 1964.

For an accessible and wide ranging treatment of spaces with constant curvatureand their groups see

[3] J. Ratcliffe, Foundations of Hyperbolic Manifolds, Graduate Texts in Mathe-matics, 91, Springer-Verlag, New York, 1994.

We shall mostly work in d-dimensional Euclidean space Ed , though sometimes

we need spherical space Sd (the unit sphere in E

d+1) , or even hyperbolic space Hd .

Although we shall often use purely geometric arguments, particularly in the planeE

2 or in ordinary space E3 , it is still useful to review some analytic methods below.

B. Euclidean Spaces

We may define Ed as a d-dimensional real vector space, equipped with an inner

product x · y (i.e. positive definite symmetric bilinear form). The norm of x ∈ Ed

is ‖x‖ = (x · x)1/2.

1. Verify the Cauchy-Schwartz inequality for x, y ∈ Ed:

| x · y | ≤ ‖x‖ ‖y‖ .

Characterize the case of equality.

2. Show that (Ed, ‖ · ‖) is a normed linear space.

We thus have a metric on Ed: the distance from x to y is ‖x − y‖.

3. Verify that Ed is then a metric space.

4. Check that Rd , the space of 1 × d row vectors, equipped with the standard

inner product x · y := xyt is a Euclidean d-space.

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Occasionally it will do to consider Ed as a linear space with origin o, say. At

other times, we require the natural affine structure on Ed . If x, y ∈ E

d , the(closed) line segment joining x to y is

[x, y] = {(1 − λ)x + λy : λ ∈ R, 0 ≤ λ ≤ 1}

Likewise, the line joining x and y is

←→

xy= {(1 − λ)x + λy : λ ∈ R} .

5. For u ∈ Ed, u ∈ [x, y] if-f

‖x − u‖ + ‖u − y‖ = ‖x − y‖ .

We allow x = y , in which case the segment or ‘line’ actually degenerates to asingle point.

If x1, ..., xr ∈ Ed , λ1, ...λr ∈ R , then

x = λ1x1 + ... + λrxr

is

• a linear combination of x1, ..., xr in any case;

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• an affine combination, if also λ1 + ... + λr = 1

• a convex combination, if furthermore all λj ≥ 0 .

A subset U ⊆ Ed is a linear (resp affine) subspace if it is closed under

arbitrary linear (resp. affine) combinations of its elements. Similarly, U is aconvex subset of E

d if closed under arbitrary convex combinations.

If U, W ⊆ Ed and λ ∈ R, then we define

U + W := {u + w : u ∈ U, w ∈ W}

U − W := {u − w : u ∈ U, w ∈ W}

λU = {λu : u ∈ U} .

In particular, if a ∈ Ed we let

U + a := U + {a} .

6. For U, W, Z ⊆ Ed , λ, η ∈ R

(U + W ) + Z = U + (W + Z) ,

λ(U + W ) = λU + λW .

When does (λ + η)U = λU + ηU ?

7. Show that U is a linear subspace of Ed

if-f U is closed under the linear operations induced from Ed.

if-f U + λU ⊆ U , ∀λ ∈ R.

8. Show that U is an affine subspace of Ed

if-f←→

xy⊆ U for all x, y ∈ U (i.e. U is ‘line closed’)if-f U is the translate L + a of a linear subspace L ⊆ E

d.

9. Consider the affine subspace U ⊆ Ed , and suppose U = L + a for linear

subspace L.

• Show that L is uniquely determined by U .

• Show that L = U − U .

• If b ∈ U , show that U = L + b .

• Show that two translates of L are either identical or disjoint.

We say naturally enough that two translates of L are parallel. More generally,affine spaces U, W are parallel if one of the direction spaces U −U , W −Wis a subspace of the other.

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10. Suppose U is an affine subspace and b ∈ U . For λ ∈ R , define

λ ∗ U := unique translate of U passing through λb .

Show that this definition is independent of b and that

λ ∗ U = λb + (U − U) = (λ − 1)b + U .

Show that λ ∗ U = λU if λ 6= 0. However,

0 ∗ U = U − U .

(P. McMullen has used this useful convention in describing realizations of apeiro-hedra.)

11. A subset U ⊆ Ed is convex if-f it contains the line segment [x, y] for all x, y ∈

U .

12. The intersection of any family of linear subspaces is a linear subspace; ditto foraffine subspaces, convex subsets.

For any subset X ⊆ Ed , we may define linear hull linX to be the intersection

of all linear subspaces containing X . The affine hull aff X and convex hull

conv X are similarly defined.

13. For any X , linX is a linear subspace; aff X is an affine subspace; convX is aconvex set.

14. Prove that linX equals the set of all linear combinations of elements of X .Prove analogous statements for affX and convX .

15. Prove

lin(linX) = linX

aff(affX) = affX

conv(convX) = convX

aff(convX) = affX

The convex hull of a finite set X in Ed is called a convex polytope.

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We now discuss dimension. A subset X ⊆ Ed is linearly dependent if there

exist x1, ..., xr ∈ X and λ1, ..., λr ∈ R , not all 0, such that

o = λ1x1 + ... + λxr .

If also λ1 + ... + λr = 0 , we say X is affinely dependent. If this does nothappen, then the set X is linearly (resp. affinely) independent.

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16. X is affinely dependent

• if-f X + a is affinely dependent, for any a ∈ Ed.

• if-f some x ∈ X is an affine combination of other xj ∈ X.

17. Let xj = (ξj1, ...ξjd) , for 0 ≤ j ≤ r.Then {x0, ..., xr} is affinely dependent

• if-f {x1 − x0, ..., xr − x0} is linearly dependent.

• if-f the matrix

1 ξ01 ... ξ0d

1 ξ11 ... ξ1d...1 ξr1 ... ξrd

has rank less than r + 1.

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Now let U be any affine subspace of Ed and let L (= U − U) be its (unique)

linear direction space. Thus U = L + a , for any a ∈ U . Let r = dim(L).

18. Every affinely independent subset of U is contained in some maximal affinelyindependent subset B . Each such B has r + 1 elements and U = aff(B) ;furthermore, every x ∈ U is uniquely expressible as an affine combination ofelements of B .

Accordingly, the set B = {b0, b1, ..., br} in the previous problem is called anaffine basis for U , and we say that dim(U) = r.

19. Interpret this in the case that {b0, b1, b2} are the vertices of a triangle in theplane E

2 (here U equals all of E2). Show that each x ∈ E

2 can be uniquelyexpressed as

x = λ0b0 + λ1b1 + λ2b2 ,

where λj ∈ R and λ0 + λ1 + λ2 = 1 .

Find a geometric meaning for the affine coordinates λj . Describe each vertex,edge and supporting line of the triangle using these affine coordinates. Thethree supporting lines divide the plane into 7 regions. Characterize each interms of affine coordinates.

20. Show that {x0, ..., xr} is an affine basis for U = L+a if-f {x1−x0, ..., xr −x0}is a (linear) basis for L = U − U .

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21. If U = aff(X) , we can choose an affine basis for U from X .

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Let L be any r-dimensional linear subspace of Ed. Recall that we can extend

any basis for L to one for Ed , and further use the Gram-Schmidt process

to produce an orthonormal basis {e1, ..., ed} for Ed , where {e1, ...er} is an

orthonormal basis for L . The orthogonal complement of L is the (d −r)−dimensional linear space L⊥ spanned by {er+1, ..., ed}. Thus E

d = L⊕L⊥.

22. Show that

L⊥ = {x ∈ Ed : x · u = 0, ∀u ∈ L} .

(Thus L⊥ can be defined independently of the basis used above.)

23. Suppose L, M are linear subspaces of Ed. Show that

(a) L ⊆ M =⇒ M⊥ ⊆ L⊥.

(b) (L⊥)⊥ = L

(c) L + M and L ∩ M are linear subspaces of Ed.

(d) (L + M)⊥ = L⊥ ∩ M⊥.

(e) (L ∩ M)⊥ = L⊥ + M⊥.

24. (a) Give a sensible definition for an orthogonal complement to an affine sub-space L.

(b) Given any affine subspace L and point x ∈ Ed , show that there exists a

unique affine subspace M through x and orthogonal to L .

25. Let L be a subset of Ed.

(a) Show that L is a 0-dimensional affine subspace of Ed if-f L = {x} for

some x ∈ Ed. (We often simply write L = x.)

(b) Show that L is a 1-dimensional affine subspace if-f L =←→

xy (a line, withx 6= y).

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A (d − 1)−dimensional (affine) subspace H is called a hyperplane.

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26. Show that any hyperplane H is defined by a linear equation

x · n = k

for some non-zero normal vector n . Indeed, show that any line of the formL = Rn + c is completely orthogonal to H.

27. Compute the (shortest) distance from the point y to (a point on) the hyperplanex · n = k.

28. Suppose b, c are distinct points in Ed. Let

H = {x ∈ Ed : ‖x − b‖ = ‖x − c‖}

(i.e. the set of all points equidistant from b, c).

Show that H is a hyperplane which passes through the midpoint of [b, c] and

which is orthogonal to the line←→

bc . Give an equation for H .

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Let X = {b0, b1, ..., br} be an affinely independent subset of Ed. The polytope

conv(X) is called an r-simplex. (We allow r < d: think of a segment or trianglein space E

d.)

29. Prove that a 1-simplex is a (proper) segment, and that a 2-simplex is a triangle.

30. Show that a d-simplex in Ed is not contained in any hyperplane H.

31. Let {b0, b1, ..., bd} be an affine basis for Ed. Suppose ‖x − bj‖ = ‖y − bj‖ for

0 ≤ j ≤ d. Show that x = y. (See problem 28.)

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C. Isometries on Ed.

An isometry on Ed is any distance preserving map

S : Ed → E

d

‖x − y‖ = ‖xS − yS‖ , ∀x, y ∈ Ed .

(Note that we shall normally write mappings on the right of their arguments:xS := (x)S , rather than S(x).)

It is easy to check that the identity map I : Ed → E

d is an isometry.

32. (a) Show that any isometry is 1 − 1.

(b) We shall show below that isometries are onto, hence invertible. Try toprove this now. (See problem 1.)

33. (a) For each fixed vector a ∈ Ed , the translation

Ta : x 7→ x + a

is an isometry.

(b) For which a does Ta = I?

(c) The central inversion

Z : x → −x , ∀x ∈ Ed

is an isometry.

(d) Any product (i.e. composition) of isometries is an isometry.

(e) Identify the isometry T−aZTa.

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We need to manufacture a richer supply of isometries.

34. Formulate a reasonable geometric definition for reflection in a hyperplane H .Verify that the following analytic definition meets your requirements.

Let H be the hyperplane x ·n = k , with normal vector n . Then the reflection

R in H is defined by

R : x 7→ x −2

n · n(x · n − k)n

In particular, if o ∈ H , then k = 0 and

R : x → x − 2x · n

n · nn .

35.8

(a) Verify that R is an isometry.

(b) Show that R fixes each point of H and interchanges the two half-spacesinto which H decomposes E

d.

(c) Show that R2 = I . Thus R is invertible and R = R−1.

(d) If k = 0 , then R fixes the origin o and considered as a linear map hasdet(R) = −1.

(e) If x 6∈ H , then H is the perpendicular bisector of the segment [x, xR].(See problem 28.)

(f) For any x0, y0 ∈ Ed , there exists a reflection mapping x0 7→ y0 (unique if

x0 6= y0).

36. Suppose ‖x0−x1‖ = ‖y0−y1‖. Then there is a product of at most 2 reflectionswhich maps both x0 → y0 and x1 → y1.

37. Suppose X = {x0, ..., xr} and Y = {y0, ..., yr} are affinely independent subsetsof E

d with ‖xi−xj‖ = ‖yi−yj‖ for all 0 ≤ i, j ≤ r. Then there is an isometryS , in fact a product of at most r + 1 reflections, which maps each xj to yj ,0 ≤ j ≤ r.

38. Suppose that an isometry S fixes each element of an affine basis B = {b0, b1, ...bd}for E

d :

bjS = bj , 0 ≤ j ≤ d .

Then S = I. (Hint: see problem 31.)

39. (a) Every isometry S on Ed is a product of at most d + 1 reflections.

(b) If B = {b0, ..., bd} and C = {c0, ..., cd} are congruent affine bases for Ed ,

i.e. if ‖bi − bj‖ = ‖ci − cj‖ for 0 ≤ i, j ≤ d , then there exists a uniqueisometry S mapping each bj → cj.

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We now know that all isometries are invertible mappings on Ed and in fact

constitute a group, which we denote Id.

The next result asserts that a ‘local’ isometry always extends to a ‘global’isometry.

40. Let X = {xj : j ∈ J} and Y = {yj : j ∈ J} be two subsets of Ed , such that

‖xj − xk‖ = ‖yj − yk‖ , for all j, k ∈ J . Then there is an isometry S on Ed

such that xjS = yj , ∀j ∈ J . Moreover S is unique if aff(X) = Ed.

(Hints: (a) Show that with no loss of generality we may assume aff(X) ⊆ aff(Y ).(b)Suppose {x0, ..., xm} and {y0, ..., ym} are affinely independent; then as in

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problem 37 we may assume xj = yj, for 0 ≤ j ≤ m. Hence show that xm+1 6∈aff{x0, ..., xm} but with ym+1 ∈ aff{y0, ...ym} would contradict problem 31.(c)Show that aff(X) = aff(Y ) and use problem 31 again.)

41. (a) If R is reflection in the hyperplane x ·n = k , and S is any isometry, thenS−1RS is also a reflection. Identify its (hyperplane) mirror.

(b) Show that reflections R1, R2 commute if-f the corresponding normals areorthogonal: n1 · n2 = 0.

42. If the isometry T maps o to b, then isometry S fixes o if-f T−1ST fixes b.

An isometry which fixes some point b is called an orthogonal transformation.Clearly, the collection of all isometries fixing b is a group, which by the previousproblem is isomorphic, indeed conjugate to, the orthogonal group

Od = {S ∈ Id : oS = o} .

43. Each S ∈ Od is a product of at most d reflections. (Hint: see problem 1; thefirst of the d + 1 possible reflections is superfluous, since oS = o .)

Consider an isometry S ∈ Id , with oS = b , say. Applying the translation T−b

we find

ST−b =: B ∈ Od .

Thus any isometryS : x 7−→ xB + b ,

can be viewed as the product of an orthogonal transformation (fixing the origin)and a translation.

44. Suppose S : x → xB + b

U : x → xC + c ,

are two such isometries.

(a) Show that B ∈ Od and b ∈ Ed are uniquely determined by S.

(b) Compute the unique orthogonal transformation and translation componentfor S−1 and for SU .

45. Show that

Id ≃ Od ⋉ Rd

(a semi-direct product).

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46. Show that every isometry S on Rd can be (uniquely) written as

S : x → xB + b

where b ∈ Rd and B is an orthogonal d × d matrix (i.e. BBt = I).

Notice that we may consider Od to be a linear group (i.e. subgroup of GL(Ed)).We shall say S : x → xB + b is direct (sense-preserving) if det(B) = +1 , oropposite (sense-reversing) if det(B) = −1.

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47. (a) The isometry S ∈ Id is direct (resp. opposite) if-f it is a product of aneven (resp. odd) number of reflections. (See problem 1.)

(b) Show that the direct isometries constitute a subgroup ofindex 2 in Id.

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An isometry S with period 2 is called an involution (and is said to be invo-lutary). Thus S2 = I but S 6= I . Examples include the central inversion Z(in o) and all reflections R .

48. Suppose S is an involutory isometry.

(a) For any a ∈ Ed , S fixes the midpoint b of the segment [a, aS].

(b) If b is the only fixed point of S then S = T−bZTb , i.e. the centralinversionin b .

(c) The central inversion Z (at the origin o) is a product of reflections in dmutually perpendicular hyperplanes through o.

(d) Some conjugate of the involutary isometry S fixes o.

49. Let S be an involutory isometry.

(a) The fixed space F = {x ∈ Ed : xS = x} is an S-invariant affine subspace

of Ed.

(b) Let b ∈ F , which we suppose has dimension d−r. Then the subspace Mthrough b , completely orthogonal to F , is an S-invariant r-dimensionalsubspace.

(c) S induces a central inversion on M .

(d) S is a product of reflections in r mutully perpendicular hyperplanes. (Seeproblem 48 (c).)

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Let us investigate now products of two reflections , say

R1 in the hyperplane H1 : x · n1 = k1 ,R2 in the hyperplane H2 : x · n2 = k2 .

50. Recall that the above hyperplanes H1 and H2 are parallel if-f one is a translateof the other (problem 9).

(a) Show that H1 is parallel to H2 if-f the normals n1, n2 are linearly depen-dent.

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(b) If H1 is parallel to H2 , show that R1R2 is a translation. The distance ofthe translation is twice the distance between the mirrors.

(c) Conversely, if Tb is any translation, show that Tb = R1R2 is product of tworeflections, in hyperplanes orthogonal to b. Given the latter requirement,show that either of R1 or R2 can be chosen arbitrarily.

51. Let R1, R2 be reflections in lines (i.e. ‘hyperplanes’) in E2, both passing

through o.

If α is the angle from the first to second mirror, show that R1R2 is a rotationthrough 2α, with centre o.

Because of this last problem, we say that the product of reflections R1, R2 intwo intersecting hyperplanes H1, H2 is a rotation. The fixed space H1 ∩ H2

is called the axis of rotation.

52. Suppose H1 and H2 are distinct, intersecting hyperplanes.

(a) Show that F = H1 ∩ H2 is a (d − 2)−dimensional affine subspace fixedby S = R1R2.

(b) Let L = lin{n1, n2} be the linear subspace spanned by the normals. Showthat L is two dimensional.

(c) Let b ∈ F . Show that M := L + b is a plane completely orthogonal to Fat b . Show that H1, H2 meet M in lines inclined at an angle α satisfying

cos α =n1 · n2

‖n1‖ ‖n2‖.

(d) Show that each plane M is S-invariant and that S acts on M as arotation through 2α.

(e) If S is a rotation with (d − 2)-dimensional axis F and angle 2α , showthat S = R1R2 , a product of reflections in hyperplanes H1, H2 inclinedat angle α , and with H1 ∩ H2 = F . Given this, show that either of H1

or H2 can be chosen arbitrarily.

The product of reflections in perpendicular mirrors is an involutory rotation,i.e. a half-turn. A glide refection, or just glide, is the product G = RTof a reflection R with a translation T whose vector b is parallel to the mirror

H for R. (Thus if R is reflection in the hyperplane H described by x · n = k ,then we in fact have n · b = 0. ).

53. Let G = RT be the glide described just above.

(a) If b = 0 , then the degenerate glide G = R is actually an ordinary reflec-tion.

(b) Show that RT = TR.

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(c) Show that a proper glide, with b 6= o , has no fixed points.

(d) Show that G = SR′ , where S is a half-turn whose axis is parallel to themirror for the reflection R′. Given these requirements, R′ can be replacedby reflection in any parallel mirror, adjusting the half-turn accordingly.

54. Any product RT of a reflection and translation is a glide. (Hint: resolve thevector a = a1 + a2 for T into components a1 orthogonal to and a2 parallel tothe mirror for R. Apply problem 50(c) to Ta1

and Ta2. )

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D. Isometries in the plane E2 and space E

3 - first approach

Here we shall classify all isometires in E2 or E

3 in a routine, but self-contained,way. For a more insightful approach, particularly useful in classifying the regularpolygons in ordinary space, see the next section (and chapter 1 of [1]). Referto problems 43 and 47, concerning the orthogonal group Od , i.e. isometries inE

d which fix the origin o .

55. (a) Show that every orthogonal transformation the plane E2 is either a reflec-

tion or rotation.

(b) Show that the orthogonal rotations in E2 form an abelian group SO(2)

isomorphic to R/Z .

56. (a) (Euler) Every direct orthogonal transfunction in E3 is a rotation.

(b) Show that SO(3) , the group of rotations fixing o in E3 , is non-abelian.

(c) Suppose that a rotation in SO(3) is represented by an orthogonal matrixB . Show how to compute the rotation angle from trace(B).

Consider now a product P = R1R2R3 of three reflections in E2 or E

3 . Suppose(o)P−1 = b and let R0 be reflection in the perpendicular bisector of [o, b]. (Ifo = b , take any mirror through o .) By problems 55(a) and 56(a), the directisometry R0P fixes o and so is a rotation, which can be factored as a productR′R of two reflections in mirrors through o . In fact, we can freely choose R′

and R to have perpendicular mirrors. Thus

R0P = R′R

P = (R0R′)R = SR,

the product of a half-turn S and reflection R.

57. In E2 , the product of three reflections is always a glide, perhaps degenerating

to a reflection (see problem 53(c).)

58. Theorem.

(a) In the plane E2 , every isometry is either a reflection, rotation, translation

or glide.

(b) Every isometry in E2 is the product of two involutory isometries.

Consider again P = SR above. In E3 , the axis for the half-turn S is a line L

(which may or may not intersect the mirror H for R. In any case, S = R4R5 ,a product of reflections in orthogonal mirrors H4, H5 , where we may freelychoose H5 ⊥ H. Thus

P = R5U = UR5

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where U = R4R is either a translation, when L is parallel to H , or a rotationwith axis M ⊥ H5 , when L meets H .

In the latter case, P is called a rotatory reflection, i.e. the product of arotation and reflection, whose mirror is perpendicular to the axis of rotation.

59. Explain why the central inversion is a rotatory reflection.

60. Summarize the above discussion: in E3 every product of three reflections is

either what or what?

A product P = R1R2R3R4 of four reflections can be similarly analyzed. Let(o)P = b , so P ′ = PT−b is a direct isometry fixing o , namely a rotation S withaxis L. Suppose b = b1 + b2 , where b2 ∈ L and b1 · b2 = 0. Thus Tb1 = R6R7 ,where R6, R7 are suitable reflections, whose mirrors are orthogonal to b1 , andwhere the mirror for R6 contains L . But then S = R5R6 for a suitablereflection R5. In short,

P = STb

= STb1Tb2

= R5R6R6R7Tb2

= (R5R7)Tb2

= S ′Tb2 ,

the product of a rotation S ′ whose axis L′ is parallel to the vector b2 for thetranslation Tb2 . Such a direct isometry is called a screw (or screw displace-

ment).

61. Theorem.

(a) Every isometry in E3 is a reflection, rotation, translation, glide, rotatory-

reflection or screw.

(b) Every isometry in E3 is a product of two involutory isometries.

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E. Isometries in the plane E2 and space E

3 - a second approach using

regular polygons

A by-product of the classification of isometries in E2 and E

3 (see problems 58and 61 in the previous section) is the observation that each such isometry is theproduct of two involutary isometries. In fact, this is true in general and can beproved with a careful analysis of the eigenspaces of a general orthogonal matrix,without first having a full classification of the various isometries (see [1, page3]).

Theorem. In Ed, with d ≥ 2, each isometry can be expressed as a product of

two involutary isometries.

62. Find a proof for this theorem.

In [1, ch. 1] Coxeter uses this theorem as a starting point for a clssification ofisometries which is keyed to the notion of a regular polygon.

We define a polygon P in Ed to be a sequence of points or vertices

. . . , a−1, a0, a1, a2, . . .

joined in successive pairs by the segments or edges

. . . , [a−1, a0], [a0, a1], [a1, a2], . . .

(say ej := [aj , aj+1]).

This list of vertices could be finite, say a0, a1, . . . , ap−1, in which case we takesubscripts to be residues (mod p). Thus when there are p distinct vertices,we obtain a p-gon with edges e0 = [a0, a1], e1 = [a1, a2], ending with the returnedge ep−1 = [ap−1, a0]. On the other hand, if the aj’s are distinct for all j ∈ Z,we obtain the (infinite) apeirogon {∞} .

Naturally, we agree that the dimension of P is determined by the dimensionof the affine hull of the vertex set. Thus a polygon is linear if it sits inside aline, planar if in a plane, and skew otherwise.

We say that P is regular if for some isometry S we have

aj = a0Sj , ∀j ∈ Z .

Hence P is regular if there is an isometry which induces the cyclic permutation

(. . . , a−1, a0, a1, a2, . . .)

(on the vertex set).

63. Typically one also demands that the base vertex a0 not be fixed by S. Why?What is p in this case?

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64. Describe the possible regular p-gons when S is an involution.

65. Suppose P is the regular polygon generated by the isometry S. Prove thatthere are involutary isometries R0, R1 which induce the following permutationson the vertex set:

R0 : . . . (a0, a−1)(a1, a−2)(a2, a−3) . . .

R1 : . . . (a1, a−1)(a2, a−2)(a3, a−3) . . .

(Hint: use problem 40).

More precisely, show that S = R0R1, at least on the affine hull of the vertexset, and that R1 fixes the base vertex a0.

66. Why does the previous problem not quite solve problem 62 ? Perhaps one canenhance this approach to produce a complete proof.

67. [1, page 4] Show that P is regular if and only if for each positive integer k wehave

‖a0 − ak‖ = ‖a1 − ak+1‖ = ‖a2 − ak+2‖ = . . . .

68. [1, page 4] A pentagon is regular if and only if it is equilateral and equiangular.

69. [1, page 4] A polygon is regular if and only if its edges are all equal and itsangles of each kind are all equal (Schoute, 1902). For example, a skew polygonin E

3 has two kinds of angles: the usual angle aj−1ajaj+1 at a vertex, and adihedral angle aj−1(ajaj+1)aj+2 at an edge ej.

70. [1, page 4] Describe a regular skew pentagon in E4.

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Since the isometry S used above is quite general, we observe that the classifica-tions of isometries and regular polygons, say in E

2 and E3, go hand in hand and

can be achieved by examining all posssible pairs of involutions R0, R1. More-over, we deduce from problem 49 that any involutary isometry is determinedby its fixed space F , and so is classified merely by the dimension of this fixedspace.

The actual possibilities for a regular polygon P in E2 or E

3 now fall out quiteneatly. We summarize the discussion in [1, §1.6 – 1.8] by a sequence of problems:

71. Suppose R1 is central inversion.

(a) Show that edges e0 = [a0, a1] and e−1 = [a−1, a0] are collinear.

(b) Show that P is the one dimensional apeirogon {∞}, consisting of infinitelymany equally spaced points on a line, joined consecutively. Thus P is linear.

(c) Show that R0, R1 act on this line by reflection ( = central inversion!), andS by translation.

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72. Suppose R0 is either reflection or central inversion.

(a) Show that the planes a0a−1a−2 and a−1a0a1 coincide, so that P must beplanar.

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Hence we may

• assume that R0 is a reflection or half-turn in this plane

• assume that R1 is a reflection in this plane (else R1 is a half-turn = centralinversion, and we are returned to the case in problem 71)

• assume further that if R0 is a reflection, then the mirrors for R0, R1 in-tersect, say at o (the case of parallel mirrors returns us again to problem71)

(b) So assume that R0, R1 are reflections whose mirrors intersect at o, andlet α := ∠a−1 o a0. Describe P when α is incommensurable with π. Whenα = 2dπ/p, for coprime integers d and p, we obtain a p-gon of density d inscribedin a circle. We denote this finite regular polygon by {p/d}. If p < 2d, we usuallywrite {p/(p − d)} instead: thus {3/2} really describes an equilateral triangle.

When is {p/d} convex?

(c) The final planar case has R0 a half-turn and and R1 a reflection, so that Sis a glide: see problem 53(d). Show that P is an infinite regular zig-zag, an-other realization of the combinatorial polygon {∞}. When does this realization‘degenerate’ into the linear apeirogon?

The above analysis covers all possibilities in the plane E2. The remaining pos-

sibilities in space E3 have R0 a half-turn and R1 either a reflection or half-turn.

73. Suppose that R0 is a half-turn with axis L and R1 a reflection with mirror Hin E

3.

(a) Suppose L is parallel to H . Show that S = R0R1 is a glide (problem 53);furthermore, any choice of a0 ∈ H again provides a zig-zag lying in the planethrough a0 orthogonal to H (so we return to problem 72(c)).

(b) Suppose L meets H , say at o. Then S = R0R1 is akin to a ‘sphericalglide’. In fact, S is a rotatory-reflection (see the discussion following problem58). The polygon P is now a ‘spherical zig-zag’, whose vertices lie alternatelyon two parallel ‘small’ circles. To see this another way, show that S = UR, theproduct of a rotation U through angle α with reflection in a mirror orthogonalto the axis of rotation.

If α = dπ/p, for coprime integers d, p, then S has period 2p. The polygon P isa skew 2p-gon, whose alternate vertices are inscribed in the two small circles.In fact, each set of alternate vertices gives a p-gon inscribed in its small circle.Further description requires some care; it is natural to distinguish two subcases:

• If p is odd and d is even (still coprime), show that P is a prismatic

polygon, whose vertices are those of a right prism with base {p/(1

2d)}, and

whose 2p edges are all lateral face diagonals of this prism.

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• In all other cases, P is an anti-prismatic polygon, whose vertices arethose of an antiprism with base {p/d}, and whose 2p edges are simply alledges of this antiprism.

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74. Suppose that R0 is a half-turn with axis L0 and R1 another half-turn with axisL1 in E

3.

(a) If L0 and L1 intersect or are parallel, the plane containing them is S-invariant, so that P is again planar (or even linear).

(b) So suppose L0 and L1 are skew lines with unique common perpendicular lineM . Show that S is a screw with axis M (see the discussion preceding problem61).

If a0 6∈ M , and L0 and L1 are not orthogonal, then P is skew and is inscribedin a helix. Hence P is called a helical polygon. Such polygons come naturallyin chiral pairs (i.e. left- and right-handed versions).

75. [1, page 6] Every finite regular skew polygon in E3 has an even number of

vertices.

76. [1, page 6] A skew pentagon in E3 cannot be both equilateral and equiangular.

77. [1, page 6] Can a skew hexagon in E3 be both equilateral and equiangular, but

not regular?

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F. Affine Irreducibility.

The upshot of the discussion around problem 44 is that every isometry S is anaffine isomorphism. Thus S maps any r-dimensional affine subspace to anotherof the same dimension. In particular, if S maps L to some translate

(L)S = L + c ,

then S permutes the members of

L̂ := {L + x : x ∈ Ed} ,

the family of all translates of L . Considering the usual way of projectifyingE

d , we might say that S fixes an (r − 1)-space L̂ at infinity. In particular,this happens if L is actually S-invariant.

78. If the isometry S maps the r-space L to a translate, then S maps any orthog-onal (d − r)-space M to a translate. Thus L̂ being invariant implies that M̂is invariant.

79. What is the space at infinity when L = {a} is a point?

80. How does any translation T act on the various spaces at infinity?

81. SupposeS : x → xB + b ,

where B ∈ Od . (Thus B is an orthogonal transformation fixing o.) Then L̂is S-invariant if–f its linear direction space L − L is B-invariant.

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In fact, suppose now that the linear subspaces L1, L2 are fixed orthogonalcomplements at o, and let

π1 : Ed → L1

π2 : Ed → L2

be the corresponding (linear) orthogonal projections. (If we really must repre-sent the situation when affine subspaces are orthogonal at a , then conjugateappropriately by the translation Ta .)

82. Observe that

z = zπ1 + zπ2 , ∀z ∈ Ed .

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83. Suppose S ∈ Id is any isometry which maps L1 (likewise L2 ) to some trans-late. Define a mapping Sj on Lj by

Sj : Lj → Lj

x → xSπj .

(a) Show that Sj is a well-defined isometry on Lj .

(b) Show that

zS = zπ1S1 + zπ2S2 , ∀z ∈ Ed .

(Thus S may be reconstituted in a natural way from the induced mappingson L1 and L2 .)

As a temporary notation, denote the isometry group on Lj by I(Lj) .

84. Suppose Γ is a subgroup of Id which fixes L̂1 (hence also L̂2 ) for the aboveorthogonal linear subspaces L1, L2 (i.e. each S ∈ Γ maps L1 to some trans-late). Show that the maps

ϕj : Γ → I(Lj)

S → Sj

are group homomorphisms.

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